Both functions $\lceil x \rceil, -\lfloor -x \rfloor$ satisfy $f(x+1)=f(x)+1$, so it is enough to check what happens when $0\le x <1$ (because any such $f$ satisfies $f(x)=f(\{x\})+\lfloor x \rfloor$, i.e. their values are determined by their behavior on $[0,1)$).

This is not a substitute for the perfectly good answers already given, but if you’re visually oriented, a picture can help make this very clear. In the figure below $x$ is a positive real number, and $n=\lfloor x\rfloor$. Multiplying by $-1$ pivots the whole picture $180^\circ$ degrees about $x=0$, something that even I, with virtually no visual imagination, can easily visualize.

Start with the relationship between $n$ and $x$: $n$ is the nearest integer to the left of $x$, so after the rotation $-n$ is the nearest integer to the right of $-x$. In other words, $-n=\lceil-x\rceil$. Rotate the relationship between $-x$ and $\lceil-x\rceil=-n$ about $0$ again, and of course you get the original relationship between $x$ and $\lfloor x\rfloor=n$ back, but you’re also multiplying by $-1$ so you also get the relationship $-\lceil-x\rceil=-(-n)=n$. In other words, $-\lceil-x\rceil=\lfloor x\rfloor$.

Of course the same reasoning works, both visually and mathematically, when $x<0$.